GALOIS GROUPS OVER NONRIGID FIELDS

WENFENG GAO, DAVID B. LEEP, JÁN MINÁČ, AND TARA L. SMITH

Dedicated to Paulo Ribenboim

Abstract. Let F be a field with char F ≠ 2. We show that F is a nonrigid field if and only if certain small 2-groups occur as Galois groups over F. These results provide new “automatic realizability” results for Galois groups over F. The groups we consider demonstrate the inequality of two particular metabelian 2-extensions of F which are unequal precisely when F is a nonrigid field. Using known results on connections between rigidity and existence of certain valuations, we obtain Galois-theoretic criteria for the existence of these valuations.

1. Introduction

Let F be a field with char F ≠ 2. The goal of this paper is to identify basic Galois groups that must occur if F is not a rigid field. If |Ḟ /Ḟ^{2}| ≥ 8 then it is known that F is not rigid if and only if a certain group (often denoted DC) of order 16 occurs as a Galois group over F ([MS]). It has also been shown ([LS], simplifying an earlier proof in [AGKM]) that F is rigid if and only if F^{(3)} = F^{{3}} (all notation defined below). In this paper we identify two groups, denoted G_{1} and G_{2}, where |G_{1}| = 32, |G_{2}| = 64, with the property that a field F is nonrigid if and only if at least one of G_{1} and G_{2} is realizable as a Galois group over F. In each case the corresponding Galois extension lies in F^{{3}} but not in F^{(3)}. Moreover, it is shown that extensions realizing G_{1} and G_{2} correspond, respectively, to extensions realizing the groups D ⋏C and D ⋏D, which lie inside F^{(3)}. This correspondence gives new nontrivial “automatic realizability” results for Galois groups. We also point out how to detect nonrigid elements in F using the groups G_{1} and G_{2}. Finally, since rigidity conditions on field elements are known to correspond to the existence of certain valuations ([W]), we can provide a Galois-theoretic interpretation of valuation-theoretic results.

The third author was partially supported by NSERC and the special Dean of Science Fund at UWO. He also gratefully acknowledges the support of MSRI in Berkeley during the fall of 1999.

2 WENFENG GAO, DAVID B. LEEP, JÁN MINÁČ, AND TARA L. SMITH

We begin in the following section with necessary notation and terminology, as well as an explanation of the extensions F^{(3)} and F^{{3}} and properties of their Galois groups. In section 3 we examine the group D ⋏ C and its associated group G_{1}, and in section 4 the groups D ⋏ D and G_{2} are considered. Section 5 gives the main theorems connecting nonrigidity to the realizability of the groups D ⋏ C, G_{1}, D ⋏ D and G_{2}, as well as the Galois-theoretic interpretation of the existence of a nonrigid element. We conclude in section 6 with a detailed analysis of the situation when |Ḟ /Ḟ^{2}| = 4. This is the case where the group DC is not realizable as a Galois group over F, and yet F may be nonrigid.

2. Notation, terminology and Preliminaries

We let Ḟ = F \{0}, the group of nonzero elements of F. If a_{1}, … a_{n} ∈ Ḟ, the notation ⟨a_{1}, …, a_{n}⟩ denotes the quadratic form a_{1}x_{1}^{2} + ⋯ + a_{n}x_{n}^{2}. If q is a quadratic form defined over F, the notation D_{F} (q) denotes the set of elements in Ḟ represented by q over F. Basic results on quadratic forms can be found in [La].

If G is a group and σ, τ ∈ G, then [σ, τ] denotes the commutator σ^{−1}τ^{−1}στ.

The notion of a rigid field and a rigid element are important in what follows. An element a ∈ Ḟ \ ± F^{2} is rigid if D_{F} (⟨1, a⟩) = Ḟ^{2} ∪ aḞ^{2} and is nonrigid otherwise. An element a ∈ Ḟ \ ± F^{2} is double rigid if a and −a are both rigid. A field F is rigid if each a ∈ Ḟ \ ± F^{2} is rigid and is nonrigid otherwise.

Lemma 2.1. The following are equivalent.

1. F is nonrigid.

2. There exists a ∈ Ḟ \F^{2} such that |D_{F} (⟨1, −a⟩)/Ḟ^{2}| ≥ 4.

3. There exists a ∈ Ḟ \ ± F^{2} such that |D_{F} (⟨1, −a⟩)/Ḟ^{2}| ≥ 4.

Proof. It is clear that (1) and (3) are equivalent and that (3) implies (2). Now assume that (2) holds but that each element in Ḟ \ ± F^{2} is rigid. Then we can assume a = −1 in (2) and that −1 ∉ F^{2}. Then there exists b ∈ Ḟ \±F^{2} such that b ∈ D_{F} (⟨1, 1⟩). Then −1 ∈ D_{F} (⟨1, −b⟩) = Ḟ^{2} ∪ −bḞ^{2}, which is a contradiction.

An element a ∈ Ḟ is basic if either a ∈ ±F^{2} or a is not double rigid. The set of all basic elements in F is denoted B(F). Thus Ḟ = B(F) ∪ {double rigid elements} is a disjoint union and B(F) = Ḟ^{2} ∪ −Ḟ^{2} if and only if F is a rigid field.

Let v : F → Γ ∪ {∞} be a valuation on F with valuation ring A, maximal ideal M, residue field k = A/M and value group Γ, where Γ is an ordered abelian group. Let U = A\M be the group of units of A. Then Γ ≅ Ḟ /U. The valuation v is called 2-henselian if v has a unique extension to the quadratic closure of F. If char F ≠ 2 and 1 + M ⊆ Ḟ^{2}, then v is 2-henselian ([W], Lemma 4.3). The valuation v is called 2-divisible if Γ is a 2-divisible group. Since Ḟ^{2}U is the set of elements in F whose valuation lies in 2Γ, it follows that v is 2-divisible if and only if Ḟ = Ḟ^{2}U.

The following theorem first appeared in [W], Theorem 4.4 (3).

Theorem 2.2. Suppose F is nonrigid. Then there exists a valuation v on F such that 1 + M ⊆ Ḟ^{2} and B(F) = Ḟ^{2}U.

Corollary 2.3. Suppose F is nonrigid. The following are equivalent.

1. There exists a valuation v on F such that 1 + M ⊆ Ḟ^{2}, B(F) = Ḟ^{2}U and v is not 2-divisible.

2. F contains a double rigid element. In particular, if v is a valuation on F such that 1 + M ⊆ Ḟ^{2} and B(F) = Ḟ^{2}U, then an element f ∈ F is double rigid if and only if v(f) ∉ 2Γ.

Proof. Since F is nonrigid, there exists a valuation v on F such that 1 + M ⊆ Ḟ^{2} and B(F) = Ḟ^{2}U. We have v is not 2-divisible if and only if Ḟ^{2}U ⊊ Ḟ, which is equivalent to B(F) ⊊ Ḟ, in other words, F contains a double rigid element.

We now define some of our main objects of study.

1. Let F^{(2)} = F (\sqrt{F}), the maximal multiquadratic extension of F. Then Gal(F^{(2)}/F) is an elementary abelian 2-group isomorphic to ∏_{I} ℤ/2ℤ where |I| = dim_{ℤ/2ℤ} Ḟ /Ḟ^{2}. We denote Gal(F^{(2)}/F) by G_{F}^{[2]}.

2. Let F^{\{3\}} = F^{(2)}(\sqrt{F^{(2)}}) = (F^{(2)})^{(2)}. Then F^{\{3\}} is a Galois extension of F and we denote Gal(F^{{3}}/F) by G_{F}^{{3}}.

3. Let F^{(3)} denote the subfield of F^{\{3\}} generated by all fields F^{(2)}(\sqrt{g}) such that g ∈ F^{(2)} and F^{(2)}(\sqrt{g})/F is a Galois extension. Denote Gal(F^{(3)}/F) by G_{F}^{[3]}.

Thus F ⊆ F^{(2)} ⊆ F^{(3)} ⊆ F^{{3}} and F^{(2)}/F, F^{(3)}/F, F^{{3}}/F are each Galois extensions. The groups G_{F}^{[3]}, G_{F}^{[2]} are quotients of G_{F}^{{3}}.

For the connection between G_{F}^{[3]} and the Witt ring of quadratic forms over F, see [MSp].

Lemma 2.4. 1. The group G_{F}^{{3}} has exponent dividing 4.

4 WENFENG GAO, DAVID B. LEEP, JÁN MINÁČ, AND TARA L. SMITH

2. For each σ, τ ∈ G_{F}^{{3}}, the commutator [σ, τ] has order dividing 2, and [σ, τ] = [τ, σ].

3. All commutators and squares in G_{F}^{{3}} commute with each other.

Proof. Let σ, τ ∈ G_{F}^{{3}}. Then σ^{2}|_{F^{(2)}} = 1 since G_{F}^{[2]} has exponent dividing 2 and [σ, τ]|_{F^{(2)}} = 1 since G_{F}^{[2]} is abelian. Thus σ^{2}, [σ, τ] ∈ Gal(F^{{3}}/F^{(2)}), a group of exponent dividing 2. Therefore σ^{4} = 1 and [σ, τ]^{2} = 1. Note that [τ, σ] = [σ, τ]^{−1} = [σ, τ] This proves (1) and (2).

For (3), we have from above that each commutator and square in G_{F}^{{3}} lies in Gal(F^{{3}}/F^{(2)}), which is an abelian group of exponent dividing 2.

Lemma 2.5. For every σ, τ ∈ G_{F}^{{3}}, the commutator [ [σ, τ], σ] commutes with both σ and τ.

Proof. In any group, we have the identity

[a, bc] = [a, c][a, b][ [a, b], c].

Let b = c and assume that a, b ∈ G_{F}^{{3}}. Then [a, b^{2}] = [ [a, b], b] by Lemma 2.4. Let a = [σ, τ] and b = σ. We have [ [σ, τ], σ^{2}] = 1 by Lemma 2.4. Therefore, [ [ [σ, τ], σ], σ] = 1. This implies [ [σ, τ], σ] commutes with σ.

Now let a = σ^{2} and b = τ. Since [σ^{2}, τ^{2}] = 1 by Lemma 2.4, we have [ [σ^{2}, τ], τ] = 1 and so [σ^{2}, τ] commutes with τ. But [σ^{2}, τ] = [ [σ, τ], σ] by the following argument. Let a = τ and b = σ in the identity to see

[σ^{2}, τ] = [τ, σ^{2}] = [ [τ, σ], σ] = [ [σ, τ], σ].

Remark 1. The results in Lemmas 2.4, 2.5 hold in any homomorphic image of G_{F}^{{3}}.

Proposition 2.6. Every square and commutator of elements in G_{F}^{[3]} lies in Z(G_{F}^{[3]}), the center of G_{F}^{[3]}.

Proof. Let σ, τ, ψ ∈ G_{F}^{[3]}. We must show σ^{2} and [σ, τ] commute with ψ. Since F^{(3)} is the compositum of all fields F^{(2)}(\sqrt{g}) where g ∈ F^{(2)} and F^{(2)}(\sqrt{g})/F is a Galois extension, it is sufficient to prove the result for elements of the group Gal(F^{(2)}(\sqrt{g})/F) whenever F^{(2)}(\sqrt{g})/F is a Galois extension. Since G_{F}^{[2]} is an abelian group of exponent dividing 2, it follows σ^{2} and [σ, τ] both lie in Gal(F^{(2)}(\sqrt{g})/F^{(2)}), a group of order dividing 2. As F^{(2)}(\sqrt{g})/F is a Galois extension, the group Gal(F^{(2)}(\sqrt{g})/F^{(2)}) is a normal subgroup of Gal(F^{(2)}(\sqrt{g})/F) of order 1 or 2 and thus lies in the center of Gal(F^{(2)}(\sqrt{g})/F). Therefore σ^{2} and [σ, τ] lie in Z(Gal(F^{(2)}(\sqrt{g})/F)).

Remark 2. In Lemma 2.4(3) and Proposition 2.6, the result for commutators is a consequence of the result for squares since the identity [σ, τ] = (σ^{−1})^{2}(στ^{−1})^{2}τ^{2} shows that a commutator is a product of three squares.

Proposition 2.7. Let G_{0} be a group with generators x, y and assume 1. x^{4} = y^{4} = 1,

2. [x, y]^{2} = [ [x, y], x]^{2} = [ [x, y], y]^{2} = 1,

3. [ [x, y], x] and [ [x, y], y] each commute with x and y. Then the following identities hold in G_{0}, and |G_{0}| ≤ 128.

(a) yx = xy[x, y],

(b) [x, y]x = x[x, y][ [x, y], x],

(c) [x, y]y = y[x, y][ [x, y], y]. If in addition one assumes

4. y^{2} = 1, then [ [x, y], y] = 1 and |G_{0}| ≤ 32.

Proof. Since [x, y] = [x, y]^{−1} = [y, x], we have yx = xy[y, x] = xy[x, y], which is (a). For (b), let z = [x, y] in the identity zx = xz[z, x] and prove (c) similarly.

We will next show that each element g ∈ G_{0} can be written in the form

g = x^{e_{1}}y^{e_{2}}[x, y]^{e_{3}}[ [x, y], x]^{e_{4}}[ [x, y], y]^{e_{5}},

where e_{1}, e_{2} ∈ {0, 1, 2, 3} and e_{3}, e_{4}, e_{5} ∈ {0, 1}. This will show |G_{0}| ≤ 4 · 4 · 2 · 2 · 2 = 128. Let g ∈ G_{0}. Since x^{4} = 1 and [ [x, y], x] commutes with x, (a) and (b) allow us to write g in the form g = x^{e_{1}}z where e_{1} ∈ {0, 1, 2, 3} and z involves factors of the type y, [x, y], [ [x, y], x]. Since y^{4} = 1 and [ [x, y], x] and [ [x, y], y] both commute with y, (c) allows us to write g = x^{e_{1}}y^{e_{2}}w where e_{2} ∈ {0, 1, 2, 3} and w involves factors of the type [x, y], [ [x, y], x], [ [x, y], y]. We now use (2) and (3) to show g has the required form.

Now assume (4) also holds. Then

[x, y]y = x^{−1}y^{−1}xy^{2} = x^{−1}yx = x^{−1}(xy[x, y]) = y[x, y]

and so [ [x, y], y] = 1. In this case we may assume e_{2} ∈ {0, 1} and e_{5} = 0, and therefore |G_{0}| ≤ 4 · 2 · 2 · 2 · 1 = 32.

Remark 3. The proof that [ [x, y], y] = 1 in (4) holds in any group where y^{2} = [x, y]^{2} = 1.

Corollary 2.8. 1. Any subgroup of G_{F}^{{3}} or any subgroup of a homomorphic image of G_{F}^{{3}} that is generated by two elements x, y satisfies (1)-(3) of Proposition 2.7.

2. Any subgroup of G_{F}^{[3]} or any subgroup of a homomorphic image of G_{F}^{[3]} that is generated by two elements x, y has order at most 32.

Proof. (1). Lemmas 2.4, 2.5 imply (1)-(3) hold in G_{F}^{{3}}, and these properties are preserved in any homomorphic image of G_{F}^{{3}}.

(2). As G_{F}^{[3]} is a homomorphic image of G_{F}^{{3}}, we know (1)-(3) hold. Since [x, y] ∈ Z(G_{F}^{[3]}) by Proposition 2.6, it follows [ [x, y], x] = [ [x, y], y] = 1. Therefore the subgroup has order at most 4·4·2·1·1 = 32.

The next proposition and corollary are needed in section 4.

Proposition 2.9. Let G be a metabelian group. If x, y, z ∈ G then

[ [x, y], z][ [y, z], x][ [z, x], y] = 1.

Proof. Since G is metabelian, there is a normal subgroup H such that H and G/H are abelian. Thus every commutator of G lies in H and any two commutators commute.

After completely expanding [ [x, y], z][ [y, z], x][ [z, x], y] and making obvious cancellations, one is left with

y^{−1}x^{−1}yxz^{−1}x^{−1}y^{−1}(xzyx^{−1}y^{−1}z^{−1})(yxzy^{−1}z^{−1}x^{−1})zxy.

The two expressions in parentheses commute since they are the commutators [x^{−1}, y^{−1}z^{−1}] and [y^{−1}, z^{−1}x^{−1}]. After transposing these expressions and cancelling the remaining terms, one is left with the identity.

Corollary 2.10. Let σ, τ, ψ ∈ G_{F}^{{3}} and assume [τ, ψ] = 1. Then [τ, [σ, ψ] ] = [ψ, [σ, τ] ].

Proof. Since G_{F}^{{3}} is metabelian, Proposition 2.9 implies

1 = [ [σ, τ], ψ][ [τ, ψ], σ][ [ψ, σ], τ] = [ [σ, τ], ψ][ [ψ, σ], τ].

Since all commutators have order dividing 2 by Lemma 2.4, the last expression equals [ψ, [σ, τ] ][τ, [σ, ψ] ] and the result follows.

3. The groups D ⋏ C and G_{1} as Galois groups

Let D denote the dihedral group of order 8 and let C denote the cyclic group of order 4. Let λ : D → ℤ/2ℤ be a group homomorphism with ker(λ) ≅ ℤ/2ℤ × ℤ/2ℤ and let η : C → ℤ/2ℤ be the unique nontrivial group homomorphism. We let D ⋏ C denote the pullback of this pair of homomorphisms. Thus D ⋏ C is the subgroup of D × C of order 16 consisting of elements (u, v) such that λ(u) = η(v). Note that the subgroup D ⋏ C projects onto both D and C with maps π_{1} : D ⋏ C → D and π_{2} : D ⋏ C → C such that λ ∘ π_{1} = η ∘ π_{2}.

Let a, b ∈ F and let E = F (\sqrt{a}, \sqrt{b}). Assume [E : F] = 4. We let D^{a,b} denote a Galois extension of F (if one exists) such that F ⊆ E ⊆ D^{a,b}, Gal(D^{a,b}/F) ≅ D, and Gal(D^{a,b}/F (\sqrt{ab})) ≅ C.

We let C^{a} denote a cyclic quartic extension of F (if one exists) such that F ⊆ F (\sqrt{a}) ⊆ C^{a}.

It is known that D^{a,b} exists if and only if the quaternion algebra (a, b)_{F} = 0 in the Brauer group Br(F), and this is equivalent to a ∈ im(N_{F (\sqrt{b})/F}) where N_{F (\sqrt{b})/F} is the norm map from F (\sqrt{b}) to F. The field C^{a} exists if and only if the quaternion algebra (a, a)_{F} = 0, which is equivalent to a ∈ im(N_{F (\sqrt{a})/F}) and also equivalent to a being a sum of two squares in F.

Now assume C^{a} and D^{a,b} exist, and let K denote a composite of C^{a} and D^{a,b}. We have F (\sqrt{a}) = C^{a} ∩ D^{a,b} since D does not admit C as a quotient group. Thus K/F is a Galois extension and [K : F] = 16.

Proposition 3.1. Gal(K/F) ≅ D ⋏ C.

Proof. Let r : Gal(K/F) ⟶ D × C be the group homomorphism defined by r(σ) = (σ|_{D^{a,b}}, σ|_{C^{a}}). Then r is injective since K = D^{a,b}C^{a}. Since F (\sqrt{a}) = C^{a} ∩ D^{a,b} and Gal(D^{a,b}/F (\sqrt{a})) ≅ ℤ/2ℤ × ℤ/2ℤ, it follows that r(σ) ∈ D ⋏ C. Since | Gal(K/F)| = 16 = |D ⋏ C|, it follows that r maps Gal(K/F) isomorphically onto D ⋏ C.

Proposition 3.2. The following statements are equivalent. 1. There is a Galois extension K/F such that Gal(K/F) ≅ D ⋏ C.

2. There exist elements a, b ∈ F such that [F (\sqrt{a}, \sqrt{b}) : F] = 4 and (a, a)_{F} = (a, b)_{F} = 0 in the Brauer group Br(F).

Proof. We have that (2) implies (1) by Proposition 3.1 since (2) implies that D^{a,b} and C^{a} exist. Now suppose (1) holds and let Gal(K/F) ≅ D ⋏ C. The kernel of λ ∘ π_{1} = η ∘ π_{2} has order 8 and thus the subfield of K corresponding to ker(λ ∘ π_{1}) is a quadratic exension of F that we denote F (\sqrt{a}). It follows that ker(π_{1}) is a subgroup of D ⋏ C of order 2 which corresponds to a subfield of K of the form D^{a,b} for some b ∈ F and ker(π_{2}) is a subgroup of D ⋏ C of order 4 which corresponds to a subfield of K of the form C^{a}. The existence of both D^{a,b} and C^{a} implies (2).

See [GSS] and [GS] for more information on the realizability of D⋏C as a Galois group.

8 WENFENG GAO, DAVID B. LEEP, JÁN MINÁČ, AND TARA L. SMITH

Corollary 3.3. Let K/F be a Galois extension with Gal(K/F) ≅ D⋏ C. Then K contains a unique quadratic extension of F that imbeds into a cyclic quartic extension contained in K/F. This quadratic extension imbeds into two different cyclic quartic extensions contained in K/F.

Proof. This follows from the observation that D⋏C has just two normal subgroups N_{1}, N_{2} with the property that (D ⋏ C)/N_{i} is a cyclic group of order 4 and that |N_{1} ∩ N_{2}| = 2. Indeed if D ⋏ C/N ≅ C, then N contains the commutator subgroup (D ⋏ C)′. Since (D × C)′ has order 2, it follows that (D⋏C)′ has order 2 and D⋏C/(D⋏C)′ ≅ C ×ℤ/2ℤ. One checks that C × ℤ/2ℤ has exactly two subgroups M of order 2 such that (C × ℤ/2ℤ)/M ≅ C. Then N is one of the two inverse images of M in D ⋏ C. Since |N_{1} ∩ N_{2}| = |(D ⋏ C)′| = 2, the subfields corresponding to N_{i} do not generate K, and so they must intersect in a quadratic extension of F.

Now we consider a group G_{1} (defined below) of order 32 and study how to contruct a Galois extension with Galois group isomorphic to G_{1}.

Definition 3.4. Let G_{1} be the group generated by two symbols x, y subject to the relations

1. x^{4} = y^{2} = 1, [x, y]^{2} = 1, [ [x, y], x]^{2} = 1.

2. [[x, y], x] commutes with x and y.

Proposition 3.5. |G_{1}| ≤ 32.

Proof. Since [ [x, y], y] = 1 by Remark 3 after Proposition 2.7, the result now follows from Proposition 2.7.

We now give a construction that will show |G_{1}| = 32. Let a, b ∈ F, let E = F (\sqrt{a}, \sqrt{b}) and assume [E : F] = 4. Assume in addition that D^{a,b} and C^{a} exist. Then a ∈ im(N_{F (\sqrt{b})/F})∩im(N_{F (\sqrt{a})/F}) and so there exist α ∈ F (\sqrt{a}) and β ∈ F (\sqrt{b}) such that N_{F (\sqrt{a})/F} (α) = N_{F (\sqrt{b})/F} (β) = a. Then [Wd], Lemma 2.14, implies there exist γ ∈ E and d ∈ F such that N_{E/F (\sqrt{a})}(γ) = αd and N_{E/F (\sqrt{b})}(γ) = βd. Let L = E(\sqrt{αd}, \sqrt{βd}, \sqrt{γ}) and K = E(\sqrt{αd}, \sqrt{βd}).

Proposition 3.6. 1. K/F is a Galois extension with Gal(K/F) ≅

D ⋏ C.

2. L/F is a Galois extension with Gal(L/F) ≅ G_{1}.

3. K ⊆ F^{(3)}, L ⊆ F^{{3}}, L ⊈ F^{(3)}.

Proof. The extensions F (\sqrt{αd})/F and E(\sqrt{βd})/F are Galois with

Gal(F (\sqrt{αd})/F) ≅ C, Gal(E(\sqrt{βd})/F) ≅ D,

because N_{F (\sqrt{a})/F} (αd) = ad^{2} and N_{F (\sqrt{b})/F} (βd) = ad^{2}. We also note that Gal(E(\sqrt{βd})/F (\sqrt{ab})) ≅ C because F (\sqrt{ab}) ⊆ F (\sqrt{ab})(\sqrt{a}) = E ⊆ E(\sqrt{βd}) and N_{E/F (\sqrt{ab})}(βd) = N_{F (\sqrt{b})/F} (βd) = ad^{2}. We see that K is the composite of F (\sqrt{αd}) and E(\sqrt{βd}), so K/F is Galois and Proposition 3.1 implies Gal(K/F) ≅ D ⋏ C.

We have K ⊆ F^{(3)} since F (\sqrt{αd}) = F (\sqrt{a})(\sqrt{αd}) ⊆ F^{(3)} and E(\sqrt{βd}) ⊆ F^{(3)}. This proves (1) and the first part of (3).

Since L = K(\sqrt{γ}), in order to show L/F is Galois, it is sufficient to show γσ(γ) ∈ K^{2} for all σ ∈ Gal(K/F). Since γ ∈ E, we need only consider σ ∈ Gal(E/F). This is clear for σ = 1, and the others follow from the calculations N_{E/F (\sqrt{a})}(γ) = αd ∈ K^{2}, N_{E/F (\sqrt{b})}(γ) = βd ∈ K^{2} and

N_{E/F (\sqrt{ab})}(γ) = \frac{N_{E/F} (γ)γ^{2}}{N_{E/F (\sqrt{a})}(γ)N_{E/F (\sqrt{b})}(γ)} = \frac{ad^{2}γ^{2}}{αdβd} ∈ K^{2}.

Therefore L/F is a Galois extension.

We have L ⊆ F^{{3}} since αd, βd, γ ∈ E ⊆ F^{(2)}. To show L ⊈ F^{(3)} it is sufficient to show F^{(2)}(\sqrt{γ})/F is not a Galois extension. There exists σ ∈ Gal(F^{(2)}/F) such that γσ(γ) = N_{E/F (\sqrt{a})}(γ) = αd. But αd ∉ (F^{(2)})^{2} since Gal(F (\sqrt{αd})/F) ≅ C, and thus F^{(2)}(\sqrt{γ})/F is not a Galois extension. This finishes the proof of (3).

We now show Gal(L/F) ≅ G_{1}. We have [L : F] = 32 since [K : F] = 16 and K ⊊ L by (3). There exist σ_{a}, σ_{b} ∈ Gal(L/F) such that σ_{a}(\sqrt{a}) = −\sqrt{a}, σ_{a}(\sqrt{b}) = \sqrt{b}, σ_{b}(\sqrt{a}) = \sqrt{a}, σ_{b}(\sqrt{b}) = −\sqrt{b} and σ_{b}(\sqrt{αd}) = \sqrt{αd}. (To see that σ_{b} exists, note that an automorphism with this property lies in Gal(K/F) and that it can be extended to Gal(L/F).) The group generated by σ_{a} and σ_{b} equals Gal(L/F), otherwise the fixed field of this subgroup would contain a quadratic extension of F, but none of \sqrt{a}, \sqrt{b}, \sqrt{ab} is fixed by both σ_{a} and σ_{b}. We will now show that σ_{a} and σ_{b} satisfy the relations in Definition 3.4. Since the group generated by σ_{a} and σ_{b} has order 32 and |G_{1}| ≤ 32, this will imply that Gal(L/F) ≅ G_{1}.

Lemmas 2.4, 2.5 and Remark 1 following these lemmas, along with the fact that σ_{a} and σ_{b} generate Gal(L/F), imply that σ_{a}^{4} = [σ_{a}, σ_{b}]^{2} = [[σ_{a}, σ_{b}], σ_{a}]^{2} = 1, and [[σ_{a}, σ_{b}], σ_{a}] ∈ Z(G_{1}). It remains to prove σ_{b}^{2} = 1.

Since we know Gal(E(\sqrt{βd})/F) ≅ D, Gal(E(\sqrt{βd})/F (\sqrt{ab})) ≅ C and σ_{b}(\sqrt{ab}) = −\sqrt{ab}, it follows that σ_{b}|_{E(\sqrt{βd})} has order 2. Thus σ_{b}|_{K} has order 2. We now show σ_{b}^{2}(\sqrt{γ}) = \sqrt{γ}. We have γσ_{b}(γ) = αd and thus \sqrt{γ}σ_{b}(\sqrt{γ}) = (−1)^{ϵ}\sqrt{αd}, where ϵ ∈ \{0, 1\}. Then σ_{b}(\sqrt{γ})σ_{b}^{2}(\sqrt{γ}) = σ_{b}((−1)^{ϵ}\sqrt{αd}) = (−1)^{ϵ}\sqrt{αd}. These equations imply σ_{b}^{2}(\sqrt{γ}) = \sqrt{γ}.

10 WENFENG GAO, DAVID B. LEEP, JÁN MINÁČ, AND TARA L. SMITH

It follows that Gal(L/F) ≅ G_{1} since the relations in Definition 3.4 hold and [L : F] = 32.

Theorem 3.7. The following statements are equivalent.

1. There is a Galois extension K/F such that Gal(K/F) ≅ D ⋏ C.

2. There is a Galois extension L/F such that Gal(L/F) ≅ G_{1}.

3. There exist a, b ∈ F such that [F (\sqrt{a}, \sqrt{b}) : F] = 4 and (a, a)_{F} = (a, b)_{F} = 0.

4. There exists a ∈ D_{F} (⟨1, 1⟩)\Ḟ^{2} such that |D_{F} (⟨1, −a⟩)| ≥ 4.

5. F is not rigid and D_{F} (⟨1, 1⟩) ⊈ F^{2} ∪ −F^{2}.

6. Either −1 ∈ F^{2} and F is not rigid or −1 ∉ F^{2} and D_{F} (⟨1, 1⟩) ⊈ F^{2} ∪ −F^{2}.

If F is formally real, then statements (1)-(6) hold if and only if F is not a Pythagorean field.

Proof. We have already seen in Proposition 3.2 and the proof of Proposition 3.6 that (1)-(3) are equivalent. Note that (2) implies (1) because D ⋏ C is a quotient of G_{1}. The equivalence of (3) and (4) is easy to check. It is clear that (5) implies (6).

(6) ⇒ (3): First assume −1 ∈ F^{2} and F is not rigid. Then there exists a ∈ F with a ∉ ±F^{2} such that b ∈ D_{F} (⟨1, a⟩) and b ∉ F^{2} ∪ aF^{2}. Then (3) holds since (a, b)_{F} = 0 and −1 ∈ F^{2} implies (a, a)_{F} = (a, −a)_{F} = 0. Now assume −1 ∉ F^{2} and D_{F} (⟨1, 1⟩) ⊈ F^{2} ∪ −F^{2}. Let a ∈ D_{F} (⟨1, 1⟩) with a ∉ ±F^{2}. Then [F (\sqrt{a}, \sqrt{−1}) : F] = 4 and (a, a)_{F} = 0. Now (3) holds with b = −a.

(3) ⇒ (5): Suppose D_{F} (⟨1, 1⟩) ⊆ F^{2} ∪ −F^{2}. We have a ∈ D_{F} (⟨1, 1⟩) since (a, a)_{F} = 0. Since a ∉ F^{2}, we have a ∈ −F^{2}. Then (−1, b)_{F} = 0 and thus b ∈ D_{F} (⟨1, 1⟩). This is a contradiction since b ∉ F^{2} ∪ aF^{2} = F^{2} ∪ −F^{2}. Therefore D_{F} (⟨1, 1⟩) ⊈ F^{2} ∪ −F^{2}. Since (a, a)_{F} = (a, b)_{F} = 0 implies a, b ∈ D_{F} (⟨1, −a⟩), we have 1, a, b, ab ∈ D_{F} (⟨1, −a⟩) and so D_{F} (⟨1, −a⟩) contains at least four square classes. Thus F is not rigid by Lemma 2.1.

Now assume F is formally real. If (6) holds, then −1 ∉ F^{2} and D_{F} (⟨1, 1⟩) ⊈ F^{2} ∪ −F^{2}. Thus F is not a Pythagorean field. If F is not a Pythagorean field, then D_{F} (⟨1, 1⟩) ⊈ F^{2} ∪ −F^{2}, since F is formally real. Thus (6) holds.

Nondyadic local fields are rigid. Every dyadic local field F satisfies conditions (1)-(6). In fact, a can be chosen as any nonsquare element in D_{F} (⟨1, 1⟩). All global fields (of characteristic different from 2) also satisfy conditions (1)-(6).

Proposition 3.8. Suppose Gal(L/F) ≅ G_{1}. Then there is a unique quadratic extension F (\sqrt{a}) of F such that

F ⊆ F (\sqrt{a}) ⊆ E ⊆ L,

for some subfield E of L where Gal(E/F) ≅ C. The element −a is not rigid.

Proof. Let K = F^{(3)} ∩ L. Then Gal(K/F) ≅ D ⋏ C. Corollary 3.3 implies there is a unique quadratic extension F (\sqrt{a}) of F in K such that F is contained in a cyclic quartic extension E lying in K. The construction in Proposition 3.1 along with Proposition 3.2 and Theorem 3.7 imply −a is not rigid.

In the notation of this section, the two cyclic quartic extensions that appear in Corollary 3.3 are easily seen to be F (\sqrt{a}, \sqrt{αd}) and F (\sqrt{a}, \sqrt{αdb}).

4. The groups D ⋏ D and G_{2} as Galois groups

Let λ_{1} : D → ℤ/2ℤ be a group homomorphism with ker(λ_{1}) ≅ ℤ/2ℤ×ℤ/2ℤ and let λ_{2} : D → ℤ/2ℤ be another group homomorphism with ker(λ_{2}) ≅ ℤ/2ℤ × ℤ/2ℤ. Let D ⋏ D denote the pullback of this pair of homomorphisms. Thus D ⋏ D is the subgroup of D × D of order 32 consisting of elements (u, v) such that λ_{1}(u) = λ_{2}(v). Note that the subgroup D ⋏ D projects onto D with maps π_{1} : D ⋏ D → D and π_{2} : D ⋏ D → D such that λ_{1} ∘ π_{1} = λ_{2} ∘ π_{2}.

Proposition 4.1. The following statements are equivalent. 1. There is a Galois extension K/F such that Gal(K/F) ≅ D ⋏ D.

2. There exist elements a, b, c ∈ F such that [F (\sqrt{a}, \sqrt{b}, \sqrt{c}) : F] = 8 and (a, b)_{F} = (a, c)_{F} = 0 in the Brauer group Br(F).

Proof. First assume that (2) holds. Then there exist Galois extensions D^{a,b} and D^{a,c} of F. Let K denote a composite of D^{a,b} and D^{a,c}. Then D^{a,b}∩D^{a,c} = F (\sqrt{a}) since the intersection is a Galois extension of F and the only Galois subextension of D^{a,b} of degree 4 over F is F (\sqrt{a}, \sqrt{b}). Therefore, K/F is a Galois extension of F with [K : F] = 32. Now a proof that is very similar to the proof of Proposition 3.1 shows that Gal(K/F) ≅ D ⋏ D.

Now assume that (1) holds and let Gal(K/F) ≅ D⋏D. The kernel of λ_{1} ∘ π_{1} = λ_{2} ∘ π_{2} has order 16 and thus the subfield of K corresponding to this kernel is a quadratic extension of F which we denote F (\sqrt{a}). Since both π_{1} and π_{2} are surjective, it follows that ker(π_{1}) and ker(π_{2}) are each normal subgroups of D ⋏ D of order 4 which correspond to subfields E_{1}, E_{2} of K which are Galois over F with Gal(E_{1}/F) ≅ D ≅

12 WENFENG GAO, DAVID B. LEEP, JÁN MINÁČ, AND TARA L. SMITH

Gal(E_{2}/F). Since ker(π_{1}) ∩ ker(π_{2}) = 1, it follows that ker(π_{1}) and ker(π_{2}) generate a subgroup of order 16. This subgroup is ker(λ_{1} ∘ π_{1}) since ker(π_{i}) ⊆ ker(λ_{1} ∘ π_{1}) = ker(λ_{2} ∘ π_{2}). Thus E_{1} ∩ E_{2} = F (\sqrt{a}). Since ker(λ_{i}) ≅ ℤ/2ℤ × ℤ/2ℤ, it follows E_{1} has the form D^{a,b} and E_{2} has the form D^{a,c} for some b, c ∈ Ḟ. Thus (a, b)_{F} = (a, c)_{F} = 0. We have [F (\sqrt{a}, \sqrt{b}, \sqrt{c}) : F] = 8 since D^{a,b} ∩ D^{a,c} = F (\sqrt{a}).

Proposition 4.2. The group D⋏D contains a unique abelian subgroup of order 16. This subgroup equals ker(λ_{1}) × ker(λ_{2}) and is isomorphic to (ℤ/2ℤ)^{4}.

Proof. We have that ker(λ_{1}) × ker(λ_{2}) is an abelian subgroup of D ⋏ D of order 16 that is isomorphic to (ℤ/2ℤ)^{4}. Now let H be any abelian subgroup of D ⋏ D of order 16. Since π_{i}(H) is an abelian subgroup of D, we have π_{i}(H) is a subgroup of order at most 4. Since H ⊆ π_{1}(H) × π_{2}(H) and H has order 16, it follows H = π_{1}(H) × π_{2}(H). Since π_{1}(H)×π_{2}(H) ⊆ D ⋏D, it follows π_{1}(H) = ker(λ_{1}) and π_{2}(H) = ker(λ_{2}).

Corollary 4.3. In the notation of Proposition 4.1,

Gal(K/F (\sqrt{a})) = ker(λ_{1} ∘ π_{1}) = ker(λ_{1}) × ker(λ_{2}) ≅ (ℤ/2ℤ)^{4}.

Proof. The first equality follows from the proof of Proposition 4.1. Since ker(λ_{1}) × ker(λ_{2}) ⊆ ker(λ_{1} ∘ π_{1}) and each has order 16, we have equality.

Now we consider a group G_{2} (defined below) of order 64 and study how to construct a Galois extension over a field F with Galois group isomorphic to G_{2}.

Definition 4.4. Let G_{2} be the group generated by three symbols x, y, z subject to the relations

1. x^{2} = y^{2} = z^{2} = 1, [x, y]^{2} = [x, z]^{2} = [y, z] = 1.

2. [y, [x, z]] = [z, [x, y]] has order dividing 2 and commutes with x, y and z.

Lemma 4.5. The following identities hold in G_{2} and |G_{2}| ≤ 64. 1. yx = xy[x, y], zx = xz[x, z].

2. [x, y] commutes with both x and y and [x, z] commutes with both x and z.

3. [x, z]y = y[x, z][y, [x, z] ] and [x, y]z = z[x, y][z, [x, y] ].

4. [x, z][x, y] = [x, y][x, z].

Proof. Statements (1)-(3) are proved as in Proposition 2.7. (See Remark 3 following the proof of Proposition 2.7.) Since [x, y] = xyxy, and [y, [x, z]] is in the center, we have

[x, z][x, y] = [x, z]xyxy = x[x, z]yxy = xy[x, z][y, [x, z] ]xy
= xy[x, z]xy[y, [x, z] ] = xyxy[x, z][y, [x, z] ]^{2}
= [x, y][x, z],

which proves (4).

To show |G_{2}| ≤ 64, it is sufficient to show that each element of G_{2} can be written in the form

x^{e_{1}}y^{e_{2}}z^{e_{3}}[x, y]^{e_{4}}[x, z]^{e_{5}}[y, [x, z] ]^{e_{6}},

where each e_{i} ∈ {0, 1}, 1 ≤ i ≤ 6. The relations defining G_{2} and statements (1)-(4) allow exactly this.

We now give a construction of a Galois extension L/F with [L : F] = 64 and Gal(L/F) ≅ G_{2}.

Assume a, b, c ∈ F such that [F (\sqrt{a}, \sqrt{b}, \sqrt{c}) : F] = 8 and (a, b)_{F} = (a, c)_{F} = 0 in the Brauer group Br(F). Then a ∈ im(N_{F (\sqrt{b})/F}) ∩ im(N_{F (\sqrt{c})/F}) and so there exist β ∈ F (\sqrt{b}) and γ ∈ F (\sqrt{c}) such that N_{F (\sqrt{b})/F} (β) = N_{F (\sqrt{c})/F} (γ) = a. Let E = F (\sqrt{b}, \sqrt{c}). Then [Wd], Lemma 2.14, implies there exist δ ∈ E and d ∈ F such that N_{E/F (\sqrt{b})}(δ) = βd and N_{E/F (\sqrt{c})}(δ) = γd.

Let E′ = E(\sqrt{a}), K = E′(\sqrt{βd}, \sqrt{γd}), and L = K(\sqrt{δ}).

Proposition 4.6. 1. K/F is a Galois extension with Gal(K/F) ≅

D ⋏ D.

2. L/F is a Galois extension with Gal(L/F) ≅ G_{2}.

3. K ⊆ F^{(3)}, L ⊆ F^{{3}}, L ⊈ F^{(3)}.

Proof. We have F (\sqrt{a}, \sqrt{b}, \sqrt{βd}) = D^{a,b} and F (\sqrt{a}, \sqrt{c}, \sqrt{γd}) = D^{a,c}. Since D^{a,b} and D^{a,c} are each Galois extensions of F, it follows that K = D^{a,b}D^{a,c} is a Galois extension of F. The reasoning in the proof of Proposition 4.1 lets us conclude Gal(K/F) ≅ D ⋏ D. This proves (1).

We have D^{a,b} and D^{a,c} contained in F^{(3)} and thus K ⊆ F^{(3)}. Since L = E′(\sqrt{βd}, \sqrt{γd}, \sqrt{δ}) and βd, γd, δ ∈ F^{(2)}, it follows L ⊆ F^{\{3\}}. If L ⊆ F^{(3)}, then F^{(2)}(\sqrt{δ})/F is a Galois extension. This implies N_{E/F (\sqrt{b})}(δ) = βd ∈ (F^{(2)})^{2}. But this is impossible since D^{a,b} ⊈ F^{(2)}. This proves (3).

Since [L : K] = 2 by (3) and [K : F] = 32 by (1), it follows [L : F] = 64. Since δ ∈ E and K/F is a Galois extension, in order to show L/F is a Galois extension, it is sufficient to show σ(δ)δ ∈ K^{2} for all σ ∈ Gal(E/F). This is accomplished as in the proof of Proposition 3.6.

14 WENFENG GAO, DAVID B. LEEP, JÁN MINÁČ, AND TARA L. SMITH

Now we begin to show Gal(L/F) ≅ G_{2}. There exist automorphisms σ_{a}, σ_{b}, σ_{c} ∈ Gal(L/F) such that σ_{a} fixes \sqrt{b}, \sqrt{c} and σ_{a}(\sqrt{a}) = −\sqrt{a}, and similarly σ_{b} fixes \sqrt{a}, \sqrt{c}, but not \sqrt{b}, and σ_{c} fixes \sqrt{a}, \sqrt{b}, but not \sqrt{c}.

We want to choose σ_{b}, σ_{c} more carefully as follows. Choose σ_{b} ∈ Gal(L/F) such that

σ_{b}|_{F (\sqrt{a},\sqrt{c},\sqrt{γd})} = 1, σ_{b}^{2}|_{F (\sqrt{a},\sqrt{b},\sqrt{βd})} = 1, \sqrt{βd} σ_{b}(\sqrt{βd}) = \sqrt{a}d.

To see this is possible, observe that Gal(F (\sqrt{a}, \sqrt{b}, \sqrt{βd})/F) ≅ D and hence there exists σ_{b} such that the second and third conditions hold. Since K = F (\sqrt{a}, \sqrt{b}, \sqrt{βd})F (\sqrt{a}, \sqrt{c}, \sqrt{γd}), Gal(K/F) ≅ D ⋏ D and σ_{b}|_{F (\sqrt{a})} = 1, the first condition can be arranged simultaneously with the second and third.

Similarly we can choose σ_{c} ∈ Gal(L/F) such that

σ_{c}|_{F (\sqrt{a},\sqrt{b},\sqrt{βd})} = 1, σ_{c}^{2}|_{F (\sqrt{a},\sqrt{c},\sqrt{γd})} = 1, \sqrt{γd} σ_{c}(\sqrt{γd}) = \sqrt{a}d.

We now show σ_{a}, σ_{b}, σ_{c} satisfy the relations in Definition 4.4. Since σ_{a} fixes each element of E and δ ∈ E, it follows that σ_{a}^{2}(\sqrt{δ}) = \sqrt{δ}. Similar reasoning shows that σ_{a}^{2} fixes E′ and K elementwise and hence σ_{a}^{2} = 1.

We have that σ_{b}^{2}|_{K} = σ_{c}^{2}|_{K} = 1. To show σ_{b}^{2} = σ_{c}^{2} = 1, it is enough to show σ_{b}^{2}(\sqrt{δ}) = \sqrt{δ} and σ_{c}^{2}(\sqrt{δ}) = \sqrt{δ}. Since

δ σ_{b}(δ) = N_{E/F (\sqrt{c})}(δ) = γd,

we have \sqrt{δ} σ_{b}(\sqrt{δ}) = (−1)^{ϵ}\sqrt{γd} and thus

σ_{b}(\sqrt{δ}) σ_{b}^{2}(\sqrt{δ}) = (−1)^{ϵ}σ_{b}(\sqrt{γd}) = (−1)^{ϵ}\sqrt{γd}.

This implies σ_{b}^{2}(\sqrt{δ}) = \sqrt{δ}. Since δ σ_{c}(δ) = N_{E/F (\sqrt{b})}(δ) = βd, we have \sqrt{δ} σ_{c}(\sqrt{δ}) = (−1)^{ϵ′} \sqrt{βd}, and a similar proof gives σ_{c}^{2}(\sqrt{δ}) = \sqrt{δ}.

We have [σ_{a}, σ_{b}]^{2} = [σ_{a}, σ_{c}]^{2} = 1 since Gal(L/F) is a homomorphic image of G_{F}^{{3}}. We have [σ_{b}, σ_{c}]|_{K} = 1 because σ_{b}σ_{c} = σ_{c}σ_{b} on each subfield F (\sqrt{a}, \sqrt{b}, \sqrt{βd}) and F (\sqrt{a}, \sqrt{c}, \sqrt{γd}). To show [σ_{b}, σ_{c}] = 1, it is enough to show σ_{b}σ_{c}(\sqrt{δ}) = σ_{c}σ_{b}(\sqrt{δ}).

The equation \sqrt{δ} σ_{b}(\sqrt{δ}) = (−1)^{ϵ}\sqrt{γd} gives

σ_{c}(\sqrt{δ}) σ_{c} σ_{b}(\sqrt{δ}) = (−1)^{ϵ}σ_{c}(\sqrt{γd}),

and the equation \sqrt{δ} σ_{c}(\sqrt{δ}) = (−1)^{ϵ′} \sqrt{βd} gives

σ_{b}(\sqrt{δ}) σ_{b} σ_{c}(\sqrt{δ}) = (−1)^{ϵ′}σ_{b}(\sqrt{βd}).

Since \sqrt{βd} σ_{b}(\sqrt{βd}) = \sqrt{γd} σ_{c}(\sqrt{γd}), some routine manipulation gives σ_{b}σ_{c}(\sqrt{δ}) = σ_{c}σ_{b}(\sqrt{δ}).

We now show [σ_{b}, [σ_{a}, σ_{c}] ] = [σ_{c}, [σ_{a}, σ_{b}] ] has order dividing 2 and commutes with σ_{a}, σ_{b}, σ_{c}. The two expressions are equal by Corollary 2.10. The order divides 2 by Lemma 2.4. We have [σ_{a}, σ_{b}] in the center of Gal(F (\sqrt{a}, \sqrt{b}, \sqrt{βd})/F) by Proposition 2.6 and thus [σ_{c}, [σ_{a}, σ_{b}] ]|_{F (\sqrt{a},\sqrt{b},\sqrt{βd})} = 1. Similarly, [σ_{a}, σ_{c}] is in the center of Gal(F (\sqrt{a}, \sqrt{c}, \sqrt{γd})/F) and thus [σ_{b}, [σ_{a}, σ_{c}] ]|_{F (\sqrt{a},\sqrt{c},\sqrt{γd})} = 1. Thus, [σ_{b}, [σ_{a}, σ_{c}] ]|_{K} = 1 and so

[σ_{b}, [σ_{a}, σ_{c}] ] ∈ Gal(L/K) ⊆ Z(Gal(L/F)).

Therefore [σ_{b}, [σ_{a}, σ_{c}] ] commutes with σ_{a}, σ_{b}, σ_{c}.

We have now shown that σ_{a}, σ_{b}, σ_{c} satisfy the relations in Definition 4.4. Next we show σ_{a}, σ_{b}, σ_{c} generate Gal(L/F).

Let H = ⟨σ_{a}, σ_{b}, σ_{c}⟩. If H ⊊ Gal(L/F), then there exists a quadratic extension F (\sqrt{e}) of F in the fixed field of H and F (\sqrt{e}) ⊆ F^{(3)} ∩ L = K. An easy check shows that no quadratic extension of F inside F (\sqrt{a}, \sqrt{b}, \sqrt{c}) is in the fixed field of H. Then F (\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{e}) ⊆ K and (ℤ/2ℤ)^{4} is a quotient of Gal(K/F) ≅ D ⋏ D by a normal subgroup N of order 2. This implies the commutator subgroup of D ⋏ D is contained in N and this is impossible since the commutator subgroup of D ⋏ D has order 4. Thus (ℤ/2ℤ)^{4} is not a quotient of D⋏D and therefore H = Gal(L/F). Since σ_{a}, σ_{b}, σ_{c} generate Gal(L/F) and satisfy the relations in Definition 4.4, it follows that Gal(L/F) ≅ G_{2}.

Theorem 4.7. The following statements are equivalent.

1. There is a Galois extension K/F such that Gal(K/F) ≅ D ⋏ D.

2. There is a Galois extension L/F such that Gal(L/F) ≅ G_{2}.

3. There exist a, b, c ∈ F such that [F (\sqrt{a}, \sqrt{b}, \sqrt{c}) : F] = 8 and (a, b)_{F} = (a, c)_{F} = 0.

4. There exist a, b, c ∈ Ḟ \Ḟ^{2} such that a, b, c are independent mod squares and b, c ∈ D_{F} (⟨1, −a⟩).

5. There exists a ∈ Ḟ \Ḟ^{2} such that either |D_{F} (⟨1, −a⟩)| ≥ 8 or both |D_{F} (⟨1, −a⟩)| = 4 and a ∉ D_{F} (⟨1, 1⟩).

Proof. (1) and (3) are equivalent by Proposition 4.1. The construction in Proposition 4.6 shows (3) implies (2). (2) implies (1) since D ⋏ D is a quotient of G_{2} (as the construction in Proposition 4.6 shows). It is straightforward to check that (3)-(5) are equivalent using the observation a ∈ D_{F} (⟨1, −a⟩) if and only if a ∈ D_{F} (⟨1, 1⟩).

In the following proposition, DD denotes the central product of two copies of the dihedral group D. That is, if g denotes the nontrivial element in the center of D, then DD ≅ (D × D)/{1, (g, g)}. Thus DD

16 WENFENG GAO, DAVID B. LEEP, JÁN MINÁČ, AND TARA L. SMITH

is obtained from the direct product D × D by identifying the centers of the two copies of D.

Proposition 4.8. Suppose Gal(L/F) ≅ G_{2}. Let K = F^{(3)} ∩ L, so that Gal(K/F) ≅ D ⋏ D. Then there is a unique quadratic extension F (\sqrt{a}) of F such that Gal(K/F (\sqrt{a})) is an abelian group of order 16. In addition the following hold.

1. There is a unique subgroup N of G_{2} of order 2 such that G_{2}/N ≅ D ⋏ D.

2. Gal(L/F (\sqrt{a})) ≅ DD.

3. Gal(L/F (\sqrt{a})) is the unique subgroup of G_{2} isomorphic to DD.

4. The element −a is nonrigid.

Proof. Using Proposition 4.2, let F (\sqrt{a}) be the quadratic extension of F that corresponds to the unique abelian subgroup of Gal(K/F) of order 16. Any extension K′ of F that lies in L with Gal(K′/F) ≅ D⋏D must lie in F^{(3)}. Thus K is uniquely determined, so this proves (1).

In the notation of Proposition 4.6, let D_{1} = ⟨σ_{b}, [σ_{a}, σ_{c}]⟩, let D_{2} = ⟨σ_{c}, [σ_{a}, σ_{b}]⟩ and let H = ⟨σ_{b}, σ_{c}, [σ_{a}, σ_{c}], [σ_{a}, σ_{b}]⟩. Note that D_{1} ≅ D_{2} ≅ D since σ_{b} and [σ_{a}, σ_{c}] have order 2 and [σ_{b}, [σ_{a}, σ_{c}] ] is central of order 2. A similar argument holds for D_{2}. We have |H| = 32, by observing the representation of elements of G_{2} at the end of the proof of Lemma 4.5. We show Gal(L/F (\sqrt{a})) ≅ D_{1}D_{2} ≅ DD. The groups D_{1} and D_{2} commute elementwise by Definition 4.4 and Lemma 4.5. Therefore the injections D_{1} ↦ H and D_{2} ↦ H induce a surjective homomorphism D_{1} × D_{2} ↦ H. The element

([σ_{b}, [σ_{a}, σ_{c}] ], [σ_{c}, [σ_{a}, σ_{b}] ])

lies in the kernel of the map D_{1} × D_{2} ↦ H. Since [σ_{b}, [σ_{a}, σ_{c}] ] and [σ_{c}, [σ_{a}, σ_{b}] ] are the nontrivial elements in the center of D_{1} and D_{2}, it follows that H is isomorphic to the central product D_{1}D_{2} ≅ DD. It is clear that H ⊆ Gal(L/F (\sqrt{a})) and therefore Gal(L/F (\sqrt{a})) = H ≅ DD. This proves (2).

Suppose E is a quadratic extension of F such that Gal(L/E) ≅ DD. Then there is a subfield M of L such that [L : M] = 2 and Gal(M/E) ≅ (ℤ/2ℤ)^{4}. (Note that D × D/({1, g} × {1, g}) ≅ (ℤ/2ℤ)^{4}.) Thus M = E(\sqrt{α_{1}}, …, \sqrt{α_{4}}), where α_{i} ∈ E. The Galois closure of E(\sqrt{α_{i}})/F lies in F^{(3)} for each i and hence the Galois closure of M/F lies in F^{(3)}. Thus M lies in F^{(3)} and it follows M = K. Thus Gal(M/F) = Gal(K/F) ≅ D ⋏ D. Since D ⋏ D contains a unique abelian subgroup of order 16, it follows E = F (\sqrt{a}) and this proves (3).

We have from before that (a, b)_{F} = (a, c)_{F} = 0 and this implies −a is not rigid by Lemma 2.1.

5. Main Theorems

Theorem 5.1. F is not a rigid field if and only if either G_{1} or G_{2} occurs as a Galois group over F.

Proof. First assume that F is not a rigid field. Then Lemma 2.1 implies there exists a ∈ Ḟ \F^{2} such that |D_{F} (⟨1, −a⟩)/Ḟ^{2}| ≥ 4. Then either the statement in Theorem 3.7(4) holds or the statement in Theorem 4.7(5) holds. Therefore, either G_{1} or G_{2} occurs as a Galois group over F.

Now assume that either G_{1} or G_{2} occurs as a Galois group over F. Then by Theorems 3.7 and 4.7, there exists an element a ∈ Ḟ \F^{2} such that D_{F} (⟨1, −a⟩) contains at least four square classes. Then F is nonrigid by Lemma 2.1.

Theorem 5.2. Let a ∈ Ḟ \(F^{2} ∪ −F^{2}). Then the element −a is not rigid in F if and only if at least one of the following two statements holds.

1. An imbedding F ⊆ F (\sqrt{a}) ⊆ E ⊆ L exists such that Gal(E/F) ≅ C and Gal(L/F) ≅ G_{1}.

2. An imbedding F ⊆ F (\sqrt{a}) ⊆ L exists such that Gal(L/F (\sqrt{a})) ≅ DD and Gal(L/F) ≅ G_{2}.

Proof. This follows from Proposition 3.8, Proposition 4.8, the construction in Proposition 3.6 and the construction in Proposition 4.6.

Corollary 5.3. Assume F is nonrigid. Let v be a valuation on F satisfying 1+M ⊆ Ḟ^{2} and B(F) = Ḟ^{2}U. Then the following statements are equivalent.

1. v is not 2-divisible.

2. There exists an element a ∈ Ḟ \ ± F^{2} such that the two imbedding problems in Theorem 5.2 have no solutions for both F (\sqrt{a}) and F (\sqrt{−a}).

Proof. If v is not 2-divisible, then Corollary 2.3 implies there exists an element a ∈ Ḟ \ ± F^{2} such that a is double rigid. Now Theorem 5.2 implies that both statements in Theorem 5.2 fail for both F (\sqrt{a}) and F (\sqrt{−a}). Conversely, if both statements in Theorem 5.2 fail for both F (\sqrt{a}) and F (\sqrt{−a}), then a and −a are rigid. Thus v is not 2-divisible by Corollary 2.3.

6. The case |Ḟ /(Ḟ)^{2}| = 4

Proposition 6.1. Assume |Ḟ /(Ḟ)^{2}| = 4.

18 WENFENG GAO, DAVID B. LEEP, JÁN MINÁČ, AND TARA L. SMITH

1. The group G_{1} occurs as a Galois group over F if and only if F is nonrigid.

2. If F is nonreal, then G_{1} occurs as a Galois group over F if and only if Br_{2}(F) = 0 (i.e., (a, b)_{F} = 0 for all a, b ∈ F).

3. If F is formally real, then G_{1} occurs as a Galois group over F if and only if F is uniquely ordered, which in this case is equivalent to F not being Pythagorean.

Proof. (1). If G_{1} occurs as a Galois group over F, then F is nonrigid by Theorem 3.7. Now assume F is nonrigid. Then Lemma 2.1 implies there exists a ∈ Ḟ \F^{2} such that |D_{F} (⟨1, −a⟩)/Ḟ^{2}| ≥ 4. Thus D_{F} (⟨1, −a⟩) = Ḟ. There exists b ∈ F such that Ḟ = {1, a, b, ab}Ḟ^{2} and so statement (3) of Theorem 3.7 holds and G_{1} occurs as a Galois group over F.

(2). Let F = F^{2} ∪ aF^{2} ∪ bF^{2} ∪ abF^{2}. Then [F (\sqrt{a}, \sqrt{b}) : F] = 4. If Br_{2}(F) = 0, then (a, a)_{F} = (a, b)_{F} = 0. Thus G_{1} occurs as a Galois group over F by Theorem 3.7.

Now assume G_{1} occurs as a Galois group over F. Then there exist a, b ∈ F such that [F (\sqrt{a}, \sqrt{b}) : F] = 4 and (a, a)_{F} = (a, b)_{F} = 0. It follows that F = F^{2} ∪ aF^{2} ∪ bF^{2} ∪ abF^{2} and thus (a, Ḟ)_{F} = 0. Suppose first that either −1 ∈ F^{2} or a ∈ −F^{2}. Then (−1, Ḟ)_{F} = 0. This implies (b, b)_{F} = (−1, b)_{F} = 0. Then (b, Ḟ)_{F} = 0 (since (b, a)_{F} = 0 and (b, b)_{F} = 0) and so (ab, Ḟ)_{F} = 0. Therefore Br_{2}(F) = 0.

Now assume −1 ∉ F^{2} and a ∉ −F^{2}. Then we can assume b = −1 (since we also have (a, ab)_{F} = 0) and so F = F^{2}∪−F^{2}∪aF^{2}∪−aF^{2}. It is clear that F^{2} ∪aF^{2} ⊆ D_{F} (⟨1, a⟩) and we have F^{2} ∪aF^{2} ⊆ D_{F} (⟨1, 1⟩) since (−1, a)_{F} = 0. Since F^{2} ∪ aF^{2} is not an ordering of F, we must have either F^{2} ∪ aF^{2} ⊊ D_{F} (⟨1, a⟩) or F^{2} ∪ aF^{2} ⊊ D_{F} (⟨1, 1⟩). In the first case we have (−a, Ḟ)_{F} = 0 and in the second case we have (−1, Ḟ)_{F} = 0. Each case implies Br_{2}(F) = 0.

(3). Since F is formally real and |Ḟ /(Ḟ)^{2}| = 4, there is an ordering such that a is positive in this ordering and F = F^{2}∪−F^{2}∪aF^{2}∪−aF^{2}. If F is not Pythagorean then a ∈ D_{F} (⟨1, 1⟩), and thus F is uniquely ordered since F^{2} ∪aF^{2} is contained in the positive cone of any ordering of F. If F is Pythagorean, then one checks that −1, a ∉ D_{F} (⟨1, −a⟩) and so F^{2} ∪ −aF^{2} is also the positive cone of an ordering. Thus F is not uniquely ordered. The result now follows from Theorem 3.7.

Proposition 6.2. Assume |Ḟ /(Ḟ)^{2}| = 4, F is formally real and not Pythagorean. Then [F^{(3)} : F] = 16 with Gal(F^{(3)}/F) ≅ D ⋏ C, and [F^{{3}} : F] = 32 with Gal(F^{{3}}/F) ≅ G_{1}.

Proof. We have F = F^{2}∪−F^{2}∪aF^{2}∪−aF^{2} and D_{F} (⟨1, 1⟩) = F^{2}∪aF^{2} where F^{2} ∪ aF^{2} is the positive cone of the unique ordering on F. Let E = F (\sqrt{−1}). Then the square class exact sequence ([La], p. 202) applied to E/F gives |Ė/(Ė)^{2}| = 4 since D_{F} (⟨1, 1⟩) contains exactly two square classes. Since F^{(2)} = E(\sqrt{a}), the square class exact sequence applied to F^{(2)}/E gives |Ḟ^{(2)}/(Ḟ^{(2)})^{2}| ≤ \frac{1}{2}|Ė/(Ė)^{2}| = 8. Therefore [F^{{3}} : F^{(2)}] = |Ḟ^{(2)}/(Ḟ^{(2)})^{2}| ≤ 8 and so [F^{{3}} : F] ≤ 32. Since G_{1} occurs as a Galois group for some extension in F^{{3}}, it follows [F^{{3}} : F] = 32 and Gal(F^{{3}}/F) ≅ G_{1}. Since D ⋏C occurs as a Galois group over F and F^{(3)} ⊊ F^{{3}}, it follows Gal(F^{(3)}/F) ≅ D ⋏ C.

Proposition 6.3. Assume |Ḟ /(Ḟ)^{2}| = 4, F is nonreal and Br_{2}(F) = 0 (so that G_{1} occurs as a Galois group over F). Then [F^{(3)} : F] = 32 and [F^{{3}} : F] = 128.

Proof. We have F^{(2)} = F (\sqrt{a}, \sqrt{b}) for some a, b ∈ F. Let E = F (\sqrt{a}). The square class exact sequence applied to E/F gives |Ė/(Ė)^{2}| = \frac{1}{2}|Ḟ /(Ḟ)^{2}| = 8, since u(F) = 2 and so binary quadratic forms over F are universal. Since F is nonreal, it is known that u(E) = 2 also and thus the square class exact sequence applied to F^{(2)}/E gives |Ḟ^{(2)}/(Ḟ^{(2)})^{2}| = \frac{1}{2} · 8^{2} = 32. Thus [F^{\{3\}} : F^{(2)}] = |Ḟ^{(2)}/(Ḟ^{(2)})^{2}| = 32 and so [F^{{3}} : F] = 128.

An argument in the proof of Proposition 3.6 shows that Gal(F^{(3)}/F) is generated by any two automorphisms σ_{a}, σ_{b} that satisfy σ_{a}(\sqrt{a}) = −\sqrt{a}, σ_{a}(\sqrt{b}) = \sqrt{b}, σ_{b}(\sqrt{a}) = \sqrt{a}, σ_{b}(\sqrt{b}) = −\sqrt{b}. Then Corollary 2.8(2) implies | Gal(F^{(3)}/F)| ≤ 32. One can check that C^{a}C^{b} ⊆ F^{(3)} and D^{a,b} ⊆ F^{(3)}. Since [C^{a}C^{b} : F] = 16, Gal(C^{a}C^{b}/F) is abelian and Gal(D^{a,b}/F) is nonabelian, it follows [F^{(3)} : F] = 32.

Generators and relations for Gal(F^{{3}}/F) and Gal(F^{(3)}/F) can be found using Proposition 2.7 and Corollary 2.8.

We now briefly consider the case when |Ḟ /(Ḟ)^{2}| = 4 and F is a rigid field. Then F^{(3)} = F^{{3}} by [LS], although when |Ḟ /(Ḟ)^{2}| = 4, this is easily obtained by direct calculation. In the proposition below, Q_{8}⋏C denotes the pullback of the system of nontrivial homomorphisms Q_{8} ↦ ℤ/2ℤ and C ↦ ℤ/2ℤ, where Q_{8} is the quaternion group of order 8. The group Q_{8} ⋏ C is also the pullback of the system D ↦ ℤ/2ℤ and C ↦ ℤ/2ℤ, where the kernel of D ↦ ℤ/2ℤ is the cyclic group of order 4. See [GSS] for more details.

Proposition 6.4. Assume |Ḟ /(Ḟ)^{2}| = 4 and F is a rigid field. 1. If F is formally real, then [F^{{3}} : F] = 8 and Gal(F^{{3}}/F) ≅ D.

2. If F is nonreal, then [F^{{3}} : F] = 16.

(a) If −1 ∈ F^{2}, then Gal(F^{{3}}/F) ≅ C × C.
(b) If −1 ∉ F^{2}, then Gal(F^{{3}}/F) ≅ Q_{8} ⋏ C.

Proof. (1). We have F is Pythagorean by Proposition 6.1 and F^{(2)} = F (\sqrt{−1}, \sqrt{a}) for some a ∈ F. Two applications of the square class

exact sequence give |Ḟ (\sqrt{−1})/Ḟ (\sqrt{−1})^{2}| = 2 and |Ḟ^{(2)}/(Ḟ^{(2)})^{2}| = 2. Thus [F^{{3}} : F^{(2)}] = 2 and [F^{{3}} : F] = 8. We have Gal(F^{{3}}/F) ≅ D since D^{a,−a} is a dihedral extension of F.

(2). When F is nonreal, two applications of the square class exact sequence give |Ḟ^{(2)}/(Ḟ^{(2)})^{2}| = 4 and thus [F^{{3}} : F] = 16. Part (a) follows since each quadratic extension of F can be imbedded in a cyclic quartic extension of F. One way to show (b) is to see that C^{−1} and D^{a,−a} are extensions of F and that Gal(D^{a,−a}/F (\sqrt{−1})) ≅ C. Then Gal(F^{{3}}/F) ≅ Q_{8} ⋏ C.

References

[AGKM] A. Adem, W. Gao, D. Karagueuzian, J. Mináč, Field theory and the cohomology of some Galois groups, accepted in J. Algebra

[G] W. Gao, Ph.D. Thesis, University of Western Ontario, (1996).

[GS] H. Grundman, T. Smith, Automatic realizability of Galois groups of order 16, Proc. AMS 124, (1996), 2631-2640.

[GSS] H. Grundman, T. Smith, J. Swallow, Groups of order 16 as Galois groups, Expo. Math. 13, (1995), 289-319.

[La] T-Y. Lam, The Algebraic Theory of Quadratic Forms, W.A Benjamin, 1980.

[LS] D. Leep, T. Smith, Multiquadratic extensions, rigid fields and Pythagorean fields, accepted in Bull. London Math. Soc.

[MS] J. Mináč, T. Smith, A characterization of C-fields via Galois groups, J. Algebra, 137, (1991), 1-11.

[MSp] J. Mináč, M. Spira, Witt rings and Galois groups, Ann. Math. 144, (1996), 35-60.

[Wd] A. Wadsworth, Merkurjev’s elementary proof of Merkurjev’s Theorem, Cont. Math. A.M.S., 55 (II), (1986), 741-776.

[W] R. Ware, Valuation rings and rigid elements in fields, Canad. J. Math, 33, (1981), 1338-1355.

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E-mail address: wgao@sprintmail.com

Department of Mathematics, University of Kentucky, Lexington, KY 40506-0027 U.S.A

E-mail address: leep@ms.uky.edu

Department of Mathematics, University of Western Ontario, London, Ontario, Canada N6A 5B7

E-mail address: minac@uwo.ca

Department of Mathematical Sciences, University of Cincinnati, Cincinnati, Ohio 45221-0025 U.S.A

E-mail address: tsmith@math.uc.edu